0
$\begingroup$

A topological space is called a sequential space if a set $A ⊂ X$ is closed if and only if together with any sequence it contains all its limits.

A topological space is called a Frechet space, if for $A \subset X$ and $x \in \overline{A}$, there is a sequential of points $A$ s.t converges to $x$.

A topological space is called a US-space provided that each convergent sequence has a unique limit.

(1) Is the topological space $\mathbb{R}$ sequential? Is there a sequential closed (or open) subset in $\mathbb{R}$?

(2) Is the topological space $\mathbb{R}$ Freshet space? or $US$ ?

(3) Is the product of sequential space, also sequential space?

$\endgroup$

1 Answer 1

2
$\begingroup$

$\Bbb R$ is a metric space, so it is automatically Fréchet and hence sequential. Its closed subsets are all sequentially closed, and its open subsets are all sequentially open. It is Hausdorff, so it is automatically $US$. This post to Dan Ma’s Topology Blog contains an example of a non-sequential product of two sequential spaces. (In fact one of the factors is even first countable, and the other is Fréchet, so they are sequential in a very strong way.)

$\endgroup$
6
  • $\begingroup$ Is $[0,\omega_1)$ a non closed countably compact in $\mathbb{R}$?If the answer is yes, so $\mathbb{R}$ is not strongly $T_B$ space. Since strongly $T_B$ space is $T_B$, it follow that $\mathbb{R}$ is not $T_B$, but it is in contradiction with $T_B$ space $\mathbb{R}$. Is it right? $\endgroup$
    – Ebi
    Commented Oct 17, 2013 at 19:58
  • $\begingroup$ @Ebi: $[0,\omega_1)$ cannot be embedded in $\Bbb R$. $\endgroup$ Commented Oct 17, 2013 at 20:00
  • $\begingroup$ could you give more explain?Thanks. $\endgroup$
    – Ebi
    Commented Oct 17, 2013 at 20:14
  • $\begingroup$ @Ebi: One way to see it is to observe that $[0,\omega_1)$ is not Lindelöf: $\{[0,\alpha):\alpha<\omega_1\}$ is an open cover of $[0,\omega_1)$ with no countable subcover. $\Bbb R$, on the other hand, is second countable, so it’s hereditarily Lindelöf: every subspace of $\Bbb R$ is Lindelöf. Thus, $\Bbb R$ cannot have a subspace homeomorphic to $[0,\omega_1)$. $\endgroup$ Commented Oct 17, 2013 at 20:16
  • $\begingroup$ In cofinite topology space (X,$\tau$),$X$ is compact and $T_1$, but is not $US$ space.But it is not true for $\mathbb{R}$, why? or even in cocountably topology on $\mathbb{R}$, space $\mathbb{R}$ is $US$ and non- Hausdorff.(I mean, generally $\mathbb{R}$ is not $US$ or Hausdorff. Is it right?) $\endgroup$
    – Ebi
    Commented Oct 18, 2013 at 8:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .